Proud Merchants
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 3556 Accepted Submission(s): 1478
Problem Description
Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are still very proud even if their nation hasn’t been so wealthy any more.
The merchants were the most typical, each of them only sold exactly one item, the price was Pi, but they would refuse to make a trade with you if your money were less than Qi, and iSea evaluated every item a value Vi.
If he had M units of money, what’s the maximum value iSea could get?
The merchants were the most typical, each of them only sold exactly one item, the price was Pi, but they would refuse to make a trade with you if your money were less than Qi, and iSea evaluated every item a value Vi.
If he had M units of money, what’s the maximum value iSea could get?
Input
There are several test cases in the input.
Each test case begin with two integers N, M (1 ≤ N ≤ 500, 1 ≤ M ≤ 5000), indicating the items’ number and the initial money.
Then N lines follow, each line contains three numbers Pi, Qi and Vi (1 ≤ Pi ≤ Qi ≤ 100, 1 ≤ Vi ≤ 1000), their meaning is in the description.
The input terminates by end of file marker.
Each test case begin with two integers N, M (1 ≤ N ≤ 500, 1 ≤ M ≤ 5000), indicating the items’ number and the initial money.
Then N lines follow, each line contains three numbers Pi, Qi and Vi (1 ≤ Pi ≤ Qi ≤ 100, 1 ≤ Vi ≤ 1000), their meaning is in the description.
The input terminates by end of file marker.
Output
For each test case, output one integer, indicating maximum value iSea could get.
Sample Input
2 10
10 15 10
5 10 5
3 10
5 10 5
3 5 6
2 7 3
Sample Output
5
11
Author
iSea @ WHU
Source
Recommend
贪心 + 01背包
只是这个贪心的思想我领悟了好久的说，我自个做了好久，却发现不管我按那个排都不合适啊，这是为何啊？
后来我就看了题解，难怪我弄得不对呢，排序竟然是按p - q排的，简单证明如下：
假设有A、B两个物品，A的花费为p1， 限制金额为q1，B的花费为p2,，限制金额为q2。那么有两种情况：
1、先A后B所需空间：p1 + q2
2、先B后A所需空间：p2 + q1
对本题而言，若想在进行一个物品的购买时他所需的前面的状态都已更新过，那么需要所需空间大的在前面，即 p1 + q2 > p2 + q1则A在前面，整理得 ：
p1 - q1 > p2 - q2
eg：对于p1 = 5, q1 = 10; p2 = 3, q2 = 5;
V1 = p1 + q2 = 10;
V2 = p2 + q1 = 13;
在进行d[10] = max(d[10], d[10 - 5] + v(第一个物品的价值））即d[5]应该已经更新过，则应先2后1。
接下来就是简单的01背包了。
#include <cstdio> #include <iostream> #include <sstream> #include <cmath> #include <cstring> #include <cstdlib> #include <string> #include <vector> #include <map> #include <set> #include <queue> #include <stack> #include <algorithm> using namespace std; #define ll long long #define _cle(m, a) memset(m, a, sizeof(m)) #define repu(i, a, b) for(int i = a; i < b; i++) #define repd(i, a, b) for(int i = b; i >= a; i--) #define sfi(n) scanf("%d", &n) #define sfl(n) scanf("%I64d", &n) #define pfi(n) printf("%d ", n) #define pfl(n) printf("%I64d ", n) #define MAXN 5005 int dp[MAXN]; struct P{ int p, q, v; bool operator < (const P& t) const { return p - q > t.p - t.q; } }b[MAXN]; int main() { int n, m; while(~scanf("%d%d", &n, &m)) { repu(i, 0, n) sfi(b[i].p), sfi(b[i].q), sfi(b[i].v); _cle(dp, 0); sort(b, b + n); repu(i, 0, n) for(int j = m; j >= b[i].q && j >= b[i].p; j--) dp[j] = max(dp[j], dp[j - b[i].p] + b[i].v); pfi(dp[m]); } return 0; }